CBSE 2025 · Region 3 · Set 2 · Q26 · 2 marks
A voltage source sends a current of $\displaystyle 2$ A to a resistor of $\displaystyle 40 \Omega$ connected across it for $\displaystyle 5$ minutes. Calculate the electrical energy supplied by the source.
Marking-scheme solution
Given \(\displaystyle \mathrm{I}=2 \mathrm{~A}, \mathrm{R}=40 \Omega, \mathrm{t}=5\) minutes \(\displaystyle =300 \mathrm{~s}\)
Electrical energy \(\displaystyle =\mathrm{I}^{2} \mathrm{Rt}\)
\[\begin{gathered}
=(2 \mathrm{~A})^{2} \times 40 \Omega \times 300 \mathrm{~s} \\
=48000 \mathrm{~J}
\end{gathered}
\]
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CBSE Class 10 Science past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.